evaluate the integral.\n int\frac{dx}{sqrt{x^{2}-289}}=square \n(type an exact answer.)
evaluate the integral.\n int\frac{dx}{sqrt{x^{2}-289}}=square \n(type an exact answer.)
Answer
Explanation:
Step1: Recall integral formula
The integral $\int\frac{dx}{\sqrt{x^{2}-a^{2}}}=\ln|x + \sqrt{x^{2}-a^{2}}|+C$, where $a^{2}=289$, so $a = 17$.
Step2: Apply the formula
Substitute $a = 17$ into the formula, we get $\int\frac{dx}{\sqrt{x^{2}-289}}=\ln|x+\sqrt{x^{2}-289}|+C$.
Answer:
$\ln|x+\sqrt{x^{2}-289}|+C$